Find the directrix of the parabola with equation $y=-0.5x^2+2x+2$












2












$begingroup$



Find the directrix of the parabola with equation
$$y=-0.5x^2+2x+2$$




I did this:



$$a=-0.5, b = 2, c = 2$$



Formula for the directrix is:



$$y=-1/(4a)$$



$$y=-1/(4cdot(-0.5))=3.5$$



This is not right:



enter image description here



What went wrong? What is the proper way to do it?






algebra-precalculus analytic-geometry conic-sections







share|cite|improve this question











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  • 1




    $begingroup$
    The right formula is $y=-1/(4a)+y_0$, where $y_0$ is the ordinate of the vertex.
    $endgroup$
    – Aretino
    Jan 8 at 11:22










  • $begingroup$
    In addition, you miscomputed $-1/(4a)$.
    $endgroup$
    – Aretino
    Jan 8 at 11:23










  • $begingroup$
    @Aretino How did you derive that formula? Is that always the one to use if I want to find the directrix of a parabola. By the way it works now when I use the formula you denoted.
    $endgroup$
    – Ryan Cameron
    Jan 8 at 11:28












  • $begingroup$
    See here: en.wikipedia.org/wiki/Parabola#As_a_graph_of_a_function
    $endgroup$
    – Aretino
    Jan 8 at 11:33
















2












$begingroup$



Find the directrix of the parabola with equation
$$y=-0.5x^2+2x+2$$




I did this:



$$a=-0.5, b = 2, c = 2$$



Formula for the directrix is:



$$y=-1/(4a)$$



$$y=-1/(4cdot(-0.5))=3.5$$



This is not right:



enter image description here



What went wrong? What is the proper way to do it?






algebra-precalculus analytic-geometry conic-sections







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The right formula is $y=-1/(4a)+y_0$, where $y_0$ is the ordinate of the vertex.
    $endgroup$
    – Aretino
    Jan 8 at 11:22










  • $begingroup$
    In addition, you miscomputed $-1/(4a)$.
    $endgroup$
    – Aretino
    Jan 8 at 11:23










  • $begingroup$
    @Aretino How did you derive that formula? Is that always the one to use if I want to find the directrix of a parabola. By the way it works now when I use the formula you denoted.
    $endgroup$
    – Ryan Cameron
    Jan 8 at 11:28












  • $begingroup$
    See here: en.wikipedia.org/wiki/Parabola#As_a_graph_of_a_function
    $endgroup$
    – Aretino
    Jan 8 at 11:33














2












2








2





$begingroup$



Find the directrix of the parabola with equation
$$y=-0.5x^2+2x+2$$




I did this:



$$a=-0.5, b = 2, c = 2$$



Formula for the directrix is:



$$y=-1/(4a)$$



$$y=-1/(4cdot(-0.5))=3.5$$



This is not right:



enter image description here



What went wrong? What is the proper way to do it?






algebra-precalculus analytic-geometry conic-sections







share|cite|improve this question











$endgroup$





Find the directrix of the parabola with equation
$$y=-0.5x^2+2x+2$$




I did this:



$$a=-0.5, b = 2, c = 2$$



Formula for the directrix is:



$$y=-1/(4a)$$



$$y=-1/(4cdot(-0.5))=3.5$$



This is not right:



enter image description here



What went wrong? What is the proper way to do it?






algebra-precalculus analytic-geometry conic-sections




algebra-precalculus analytic-geometry conic-sections






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 12:10









Blue

49.7k870158




49.7k870158










asked Jan 8 at 11:19









Ryan CameronRyan Cameron

697




697








  • 1




    $begingroup$
    The right formula is $y=-1/(4a)+y_0$, where $y_0$ is the ordinate of the vertex.
    $endgroup$
    – Aretino
    Jan 8 at 11:22










  • $begingroup$
    In addition, you miscomputed $-1/(4a)$.
    $endgroup$
    – Aretino
    Jan 8 at 11:23










  • $begingroup$
    @Aretino How did you derive that formula? Is that always the one to use if I want to find the directrix of a parabola. By the way it works now when I use the formula you denoted.
    $endgroup$
    – Ryan Cameron
    Jan 8 at 11:28












  • $begingroup$
    See here: en.wikipedia.org/wiki/Parabola#As_a_graph_of_a_function
    $endgroup$
    – Aretino
    Jan 8 at 11:33














  • 1




    $begingroup$
    The right formula is $y=-1/(4a)+y_0$, where $y_0$ is the ordinate of the vertex.
    $endgroup$
    – Aretino
    Jan 8 at 11:22










  • $begingroup$
    In addition, you miscomputed $-1/(4a)$.
    $endgroup$
    – Aretino
    Jan 8 at 11:23










  • $begingroup$
    @Aretino How did you derive that formula? Is that always the one to use if I want to find the directrix of a parabola. By the way it works now when I use the formula you denoted.
    $endgroup$
    – Ryan Cameron
    Jan 8 at 11:28












  • $begingroup$
    See here: en.wikipedia.org/wiki/Parabola#As_a_graph_of_a_function
    $endgroup$
    – Aretino
    Jan 8 at 11:33








1




1




$begingroup$
The right formula is $y=-1/(4a)+y_0$, where $y_0$ is the ordinate of the vertex.
$endgroup$
– Aretino
Jan 8 at 11:22




$begingroup$
The right formula is $y=-1/(4a)+y_0$, where $y_0$ is the ordinate of the vertex.
$endgroup$
– Aretino
Jan 8 at 11:22












$begingroup$
In addition, you miscomputed $-1/(4a)$.
$endgroup$
– Aretino
Jan 8 at 11:23




$begingroup$
In addition, you miscomputed $-1/(4a)$.
$endgroup$
– Aretino
Jan 8 at 11:23












$begingroup$
@Aretino How did you derive that formula? Is that always the one to use if I want to find the directrix of a parabola. By the way it works now when I use the formula you denoted.
$endgroup$
– Ryan Cameron
Jan 8 at 11:28






$begingroup$
@Aretino How did you derive that formula? Is that always the one to use if I want to find the directrix of a parabola. By the way it works now when I use the formula you denoted.
$endgroup$
– Ryan Cameron
Jan 8 at 11:28














$begingroup$
See here: en.wikipedia.org/wiki/Parabola#As_a_graph_of_a_function
$endgroup$
– Aretino
Jan 8 at 11:33




$begingroup$
See here: en.wikipedia.org/wiki/Parabola#As_a_graph_of_a_function
$endgroup$
– Aretino
Jan 8 at 11:33










1 Answer
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$begingroup$

Complete the square $$begin{aligned} y&=-0.5x^2+2x+2\ &=-0.5(x^2-4x)+2\ &=-0.5(x^2-4x+4-4)+2\ &=-0.5[(x-2)^2-4]+2\&=-0.5(x-2)^2+4end{aligned}$$
or equivalently $$y-4=-0.5(x-2)^2.$$
The directrix is given by $$y-4=-frac{1}{4a}quad text{with}; a=-0.5$$
or $$y=4+frac{1}{2}$$






share|cite|improve this answer









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    $begingroup$

    Complete the square $$begin{aligned} y&=-0.5x^2+2x+2\ &=-0.5(x^2-4x)+2\ &=-0.5(x^2-4x+4-4)+2\ &=-0.5[(x-2)^2-4]+2\&=-0.5(x-2)^2+4end{aligned}$$
    or equivalently $$y-4=-0.5(x-2)^2.$$
    The directrix is given by $$y-4=-frac{1}{4a}quad text{with}; a=-0.5$$
    or $$y=4+frac{1}{2}$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Complete the square $$begin{aligned} y&=-0.5x^2+2x+2\ &=-0.5(x^2-4x)+2\ &=-0.5(x^2-4x+4-4)+2\ &=-0.5[(x-2)^2-4]+2\&=-0.5(x-2)^2+4end{aligned}$$
      or equivalently $$y-4=-0.5(x-2)^2.$$
      The directrix is given by $$y-4=-frac{1}{4a}quad text{with}; a=-0.5$$
      or $$y=4+frac{1}{2}$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Complete the square $$begin{aligned} y&=-0.5x^2+2x+2\ &=-0.5(x^2-4x)+2\ &=-0.5(x^2-4x+4-4)+2\ &=-0.5[(x-2)^2-4]+2\&=-0.5(x-2)^2+4end{aligned}$$
        or equivalently $$y-4=-0.5(x-2)^2.$$
        The directrix is given by $$y-4=-frac{1}{4a}quad text{with}; a=-0.5$$
        or $$y=4+frac{1}{2}$$






        share|cite|improve this answer









        $endgroup$



        Complete the square $$begin{aligned} y&=-0.5x^2+2x+2\ &=-0.5(x^2-4x)+2\ &=-0.5(x^2-4x+4-4)+2\ &=-0.5[(x-2)^2-4]+2\&=-0.5(x-2)^2+4end{aligned}$$
        or equivalently $$y-4=-0.5(x-2)^2.$$
        The directrix is given by $$y-4=-frac{1}{4a}quad text{with}; a=-0.5$$
        or $$y=4+frac{1}{2}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 9 at 10:32









        user376343user376343

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        3,9834829






























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